If a ball is thrown vertically upward with a velocity of 96 ft/s, then its height after t seconds is

s = -16t^2 + 96t. What is the velocity of the ball when it is 128 ft above the ground on its way up?

well, v = at

so, what is t when

-16t^2 + 96t = 128?
Pick the lowest value, since you want the ball on its way up.

Now plug that in to find v at that time.

To find the velocity of the ball when it is 128 ft above the ground on its way up, we need to determine the time at which the ball reaches that height. We can then use this time to find the velocity.

First, we can set up the equation for the height of the ball as a function of time:
s = -16t^2 + 96t

Since we are looking for the time when the ball is 128 ft above the ground, we can set s = 128 and solve for t:
128 = -16t^2 + 96t

Rearranging the equation:
16t^2 - 96t + 128 = 0

Next, we can solve this quadratic equation for t. We can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)

In this case, a = 16, b = -96, and c = 128. Substituting these values into the quadratic formula:
t = (-(-96) ± √((-96)^2 - 4 * 16 * 128)) / (2 * 16)

Simplifying:
t = (96 ± √(9216 - 8192)) / 32
t = (96 ± √1024) / 32
t = (96 ± 32) / 32

There are two possible values for t:
1) t = (96 + 32) / 32 = 128 / 32 = 4
2) t = (96 - 32) / 32 = 64 / 32 = 2

Since we are interested in the time when the ball is on its way up, the appropriate time is t = 2 seconds.

Finally, we can find the velocity at t = 2 seconds by taking the derivative of the height function with respect to time:
v = ds/dt

Given the height function s = -16t^2 + 96t, we can differentiate it to find the velocity function:
v = d(-16t^2 + 96t)/dt

Differentiating:
v = -32t + 96

Substituting t = 2:
v = -32(2) + 96 = -64 + 96 = 32 ft/s

Therefore, the velocity of the ball when it is 128 ft above the ground on its way up is 32 ft/s.

To find the velocity of the ball when it is 128 ft above the ground on its way up, we need to first find the time at which the ball reaches a height of 128 ft.

Given that the height of the ball is given by the equation s = -16t^2 + 96t, we can set s equal to 128 and solve for t:

-16t^2 + 96t = 128

Rearranging the equation, we get:

-16t^2 + 96t - 128 = 0

To solve this quadratic equation, we can either factor it or use the quadratic formula. Factoring might not be possible here, so we'll use the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / (2a)

For our equation, a = -16, b = 96, and c = -128. Plugging these values into the formula, we get:

t = (-96 ± √(96^2 - 4 * -16 * -128)) / (2 * -16)

Simplifying the equation further:

t = (-96 ± √(9216 - 8192)) / (-32)

t = (-96 ± √1024) / (-32)

Now, we'll find the two possible values of t:

t1 = (-96 + √1024) / (-32)

t1 = (-96 + 32) / (-32)

t1 = (-64) / (-32)

t1 = 2

t2 = (-96 - √1024) / (-32)

t2 = (-96 - 32) / (-32)

t2 = (-128) / (-32)

t2 = 4

So, the ball reaches a height of 128 ft after 2 seconds and again after 4 seconds.

Next, we'll find the velocities at these two times.

To find the velocity, we need to take the derivative of the height equation, which gives us the expression for velocity:

v = ds/dt

Given that s = -16t^2 + 96t, we can differentiate with respect to t:

v = d/dt (-16t^2 + 96t)

v = -32t + 96

Now, we'll plug in the values of t to find the velocities:

When t = 2 seconds:
v1 = -32(2) + 96
v1 = -64 + 96
v1 = 32 ft/s (velocity when the ball is 128 ft above the ground on its way up)

When t = 4 seconds:
v2 = -32(4) + 96
v2 = -128 + 96
v2 = -32 ft/s (velocity when the ball is 128 ft above the ground on its way down)

So, the velocity of the ball when it is 128 ft above the ground on its way up is 32 ft/s.